# Solve the inequality for real $x, 3(2-x) \geq 2(1-x)$

$\begin{array}{1 1} (A)\;(-6,\infty)\\(B)\;(-\infty,4]\\(C)\;(-4,\infty]\\(D)\; (-9,\infty)\end{array}$

Toolbox:
• Same Quantity can be added (a subtracted ) to (from ) both sides of the inequality with out changing the sign of the in equality.
• Same positive quantities can be multiplied or divided to both side of the in equality with out changing the sign of the inequality.
• If same negative quantity is multiplied or divided to both sides of the inequality is reversed i.e $'>'$ sign changes to $'<'$ and $'<'$ changes $'>'$ .
Step 1:
The given inequality
$3(2-x) \geq 2(1-x)$
=> $6-3x \geq 2-2x$
adding $2x$ on both sides of inequality.
=> $6-3x +2x \geq 2-2x+2x$
=> $6-x \leq 2$
adding $-6$ on both sides of inequality
=> $6-x -6 \geq 2-6$
$-x \geq -4$
Multiplying by a negative number -1 on both sides
$x \leq 4$
Step 2:
All real number less than and equal to 4 satisfy the given inequality .
The solution set is $(-\infty, 4]$
Hence B is the correct answer.